To prove that ctg a + ctg b = sin(b+a) / sin a * sin b, we can start by expressing the left side in terms of sine and cosine functions.

We know that cotangent (ctg) is the reciprocal of tangent, so ctg a = 1/tan a and ctg b = 1/tan b.

Using the trigonometric identity tan x = sin x / cos x, we can rewrite ctg a and ctg b as:

ctg a = 1 / tan a = cos a / sin a
ctg b = 1 / tan b = cos b / sin b

Now, let's add ctg a and ctg b:

ctg a + ctg b = cos a / sin a + cos b / sin b

By combining the fractions, we get:

ctg a + ctg b = (cos a sin b + cos b sin a) / (sin a * sin b)

Next, we can use the sum-to-product trigonometric identity sin(b+a) = sin a cos b + cos a sin b to rewrite the numerator.

Therefore, sin(b+a) = sin a cos b + cos a sin b

Substitute this into the equation:

ctg a + ctg b = sin(b+a) / sin a * sin b

Thus, the required equality is proved.